What is the value of 6t such that volume contained inside the planes `sqrt(1-t^(2))x+tz=2sqrt(1-t^(2))`
`z=0,x=2+(tsqrt(4t^(2)-5t+2))/(sqrt(12)(1-t^(2))^((1)/(4)))` and `|y|=2` is maximum.

Answer

B

Explanation

All the give planes are at rigt angles to `xz` plane so a cross section parallel to zx plane will be same every where. So, volume will be maximum when are of triangular cross section is maximum. When we cut these planes by the plane `y=0`, the Delta obtained is ABC, where
`AB=(tsqrt(4t^(2)-5t+2))/(sqrt(12)(1-t^(2))^((1)/(4)))`
Let `CAB=theta=` angle between first plane and `xy`- plane the `costheta=t`
Area of `DeltaABC=Delta=(1)/(2)(AB)^(2)tantheta`
`Delta=(1)/(2)(t^(2)(4t^(2)-5t+2))/(12(1-t^(2))^((1)/(2)))xx(sqrt(1-t^(2)))/(t)=(1)/(2)(4t^(3)-5t^(2)+2t)/(12)`
For example `(dDelta)/(dt)=0implies12t^(2)-10t+2=0`
`impliest=(10+-sqrt(100-96))/(24)=(1)/(2)` or `(1)/(3)`
Now `(d^(2)Delta)/(dt^(2))=24t-10`
So `Delta` is maximum at `t=(1)/(3)` and minimum at `t=(1)/(2)`
`becauset=(1)/(3)`